What is the general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi ?

Q: What is the general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi ?

The general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi is x= pi/2

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3sin^2(x)+sin(x)-4=0,0<= x<= 2pi

Solution

3 sin^{2} (x) + sin (x) - 4 = 0, 0 \leq x \leq 2 π

Solution

x = \frac{π}{2}

Degrees

x = 9 0^{\circ}

Solution steps

3 sin^{2} (x) + sin (x) - 4 = 0, 0 \leq x \leq 2 π

Solve by substitution

3 sin^{2} (x) + sin (x) - 4 = 0

Let:

sin (x) = u

3 u^{2} + u - 4 = 0

3 u^{2} + u - 4 = 0 : u = 1, u = - \frac{4}{3}

3 u^{2} + u - 4 = 0

Solve with the quadratic formula

3 u^{2} + u - 4 = 0

Quadratic Equation Formula:

For

a = 3, b = 1, c = - 4

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 4) = 7

1^{2} - 4 \cdot 3 (- 4)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 4)

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 1 + 7}{2 \cdot 3}, u_{2} = \frac{- 1 - 7}{2 \cdot 3}

u = \frac{- 1 + 7}{2 \cdot 3} : 1

\frac{- 1 + 7}{2 \cdot 3}

Add/Subtract the numbers:

- 1 + 7 = 6

= \frac{6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 1 - 7}{2 \cdot 3} : - \frac{4}{3}

\frac{- 1 - 7}{2 \cdot 3}

Subtract the numbers:

- 1 - 7 = - 8

= \frac{- 8}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 8}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{6}

Cancel the common factor:

2

= - \frac{4}{3}

The solutions to the quadratic equation are:

u = 1, u = - \frac{4}{3}

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - \frac{4}{3}

sin (x) = 1, sin (x) = - \frac{4}{3}

sin (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{2}

sin (x) = 1, 0 \leq x \leq 2 π

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}

sin (x) = - \frac{4}{3}, 0 \leq x \leq 2 π :

No Solution

sin (x) = - \frac{4}{3}, 0 \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi ?
The general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi is x= pi/2